Is the Cartesian square of the set of irrational numbers path connected?

Let $X=\mathbb{R}\setminus \mathbb{Q}$. Is $X\times X$ path-connected?

I don't know where to start I think we need some number theory knowledge.

• I don't believe it's path connected. If you just think about just $X = \mathbb{R} \backslash \mathbb{Q}$, there are gaps at all the rationals, so there conceivably could be points that are not connected by a continuous path since $\mathbb{Q}$ is dense in $\mathbb{R}$. Of course, this is not rigorous but it should give some intuition behind the answer. – Kevin Sheng Nov 19 '15 at 5:12
• @B.Pasternak $X\times X$ is not the same as that space. If you remove rational pairs from $\mathbb R^2$ you are still left with points that have a single rational coordinate. – John Douma Nov 19 '15 at 5:19
• This space is disconnected because all rational coordinates have been removed. The two open half planes separated by the $y$ axis are also closed in this space. Therefore, the space is not even connected so it cannot be path connected. – John Douma Nov 19 '15 at 5:21
• @Kevin: If $X \times X$ is path connected, pick a path from $(x,y)$ to $(x',y)$. Project onto the first coordinate. You now have a path from $x$ to $x'$. – user98602 Nov 19 '15 at 5:24
• @JohnDouma I misread, and you are right. – B. Pasternak Nov 19 '15 at 5:32

This space is not the same as $\mathbb R^2$ \ $\mathbb Q^2$.
$\mathbb R^2$ \ $\mathbb Q^2$ is path connected and if you search this site you will find a couple of nice proofs of that fact.
In this space, we remove the rational numbers before taking the product. Consider the partition of $X\times X$ into two half planes: $\{(x,y): x\lt 0\}$ and $\{(x,y): x\gt 0\}$. Each of these sets is open and since they partition the space, they are each the complement of an open set so they are also closed.
Therefore, $X\times X$ is disconnected and so it cannot be path connected.